It is true that there is a universal constant $c$ such that $\mathbb E[f(Z)]\le c\mathbb E[f(Y)]$. The best (smallest) value for $c$ that I can prove now is $c=2.187...$, but expect that it holds for $c=2$.

Let's do this in several steps.

**1) Reduce to unit variance Rademacher sums**

Conditioning on the absolute values of $X_i$ reduces to the case where $X_i=\pm a_i$ for constants $a_i$. If we can prove the result for such case, then letting $\mathcal G$ be the $\sigma$-algebra generated by the absolute values,
$$
f(Z)\le c\mathbb E[f(Y)\vert\mathcal G]
$$
and, taking expectations gives the result. It is enough to suppose that the variance $\mathbb E[Y^2]=\lVert a\rVert_2^2$ is equal to 1, since we can absorb any multiplicative factor into $f$.

**2) Reduce to f(Y)=min(Y,u)**

As $f$ is concave with $f(0)=0$,
$$
f(x)=\int_0^\infty\min(x,u)(-f''(u))\,du + xf'(\infty)
$$
over $x\ge0$. Assuming twice differentiability, this can be proved with integration by parts, although it holds in the general case where the second derivative is in the measure theoretic sense.
Putting $x=Y$ into this and taking expected values, commuting expectation with integral sign reduces the claim to the cases for $\mathbb E[\min(Y,u)]$ and $\mathbb E[Y]$ (the latter is just the limit of $\mathbb E[\min(Y,u)]$ as $u\to\infty$ anyway).

**3) Reduce to f(Y)=min(Y,1)**

We have already reduced to $f(Y)=\min(Y,u)$ and $\lVert a\rVert_2=1$ so that the inequality is
$$
c\mathbb E[\min(Y,u)]\ge f(Z)=\min(1,u).
$$
As the left hand side is increasing in $u$ and the right is just 1 when $u\ge1$, proving the $u=1$ case proves for all $u\ge1$.

When $0 < u\le1$ the right hand side is $u$ so, dividing through by this the inequality is
$$
c\mathbb E[\min(Y/u,1)]\ge1
$$
and, as the left hand side is decreasing in $u$, proving it for $u =1$ proves for all $u\le1$. So, all that remains is to show the $u=1$ case
$$
c\mathbb E[\min(Y,1)]\ge1.
$$

**4) Prove the reduced case**

$$
\mathbb E[\min(Y,1)]=\mathbb E[Y]-\mathbb E[\max(Y-1,0)]
$$
The optimal Khintchine lower bound for $\mathbb E[Y]$ is $1/\sqrt2$.
Using $y-1\le y^2/4$ then $\mathbb E[\max(Y-1,0)]$ is bounded above by $\mathbb E[Y^2/4]=1/4$ (as we reduced to the unit variance case). So,
$$
\mathbb E[\min(Y,1)]\ge1/\sqrt2-1/4.
$$
Hence, your inequality holds with
$$
c=\left(1/\sqrt2-1/4\right)^{-1}=2.187...
$$


**Note**

The conjectured optimal bound $c=2$ is obtained when $X_1=\pm1/\sqrt2$, $X_2=\pm1/\sqrt2$ and $X_i=0$ for $i > 2$. The optimal Khintchine lower bound I used for $\mathbb E[Y]$ is also obtained in the same case. As they obtain the optimal bounds simultaneously, to get a better bound, we should be able to concentrate on the upper bound for $\mathbb E[\max(Y-1,0)]$.
Specifically, $c=2$ would follow from
$$
\mathbb E[\max(Y-1,0)]\le1/\sqrt2-1/2=0.2071...
$$
I don't have a proof of this, but it seems plausible.